C3 January 2010 Q9
9.
| Scheme | Marks |
|---|---|
| \(\ln(3x - 7) = 5\) \(\mathrm{e}^{\ln(3x - 7)} = \mathrm{e}^5\) | M1 |
| \(3x - 7 = \mathrm{e}^5 \Rightarrow x = \dfrac{\mathrm{e}^5 + 7}{3}\ \{= 51.804\ldots\}\) | dM1 A1 |
| (3) |
Notes
M1: Takes e of both sides of the equation. This can be implied by \(3x - 7 = \mathrm{e}^5\).
dM1: Then rearranges to make \(x\) the subject.
A1: Exact answer of \(\dfrac{\mathrm{e}^5 + 7}{3}\).
| Scheme | Marks |
|---|---|
| \(3^x\mathrm{e}^{7x + 2} = 15\) | |
| \(\ln\left(3^x\mathrm{e}^{7x + 2}\right) = \ln 15\) | M1 |
| \(\ln 3^x + \ln\mathrm{e}^{7x + 2} = \ln 15\) | M1 |
| \(x\ln 3 + 7x + 2 = \ln 15\) | A1 oe |
| \(x(\ln 3 + 7) = -2 + \ln 15\) | ddM1 |
| \(x = \dfrac{-2 + \ln 15}{7 + \ln 3}\ \{= 0.0874\ldots\}\) | A1 oe |
| (5) |
Notes
M1: Takes ln (or logs) of both sides of the equation.
M1: Applies the addition law of logarithms.
A1 oe: \(x\ln 3 + 7x + 2 = \ln 15\)
ddM1: Factorising out at least two \(x\) terms on one side and collecting number terms on the other side.
A1 oe: Exact answer of \(\dfrac{-2 + \ln 15}{7 + \ln 3}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \mathrm{e}^{2x} + 3,\ x \in \mathbb{R}\) | |
| \(y = \mathrm{e}^{2x} + 3 \Rightarrow y - 3 = \mathrm{e}^{2x}\) \(\Rightarrow \ln(y - 3) = 2x\) \(\Rightarrow \tfrac{1}{2}\ln(y - 3) = x\) | M1 M1 |
| Hence \(\mathrm{f}^{-1}(x) = \underline{\tfrac{1}{2}\ln(x - 3)}\) | A1 cao |
| \(\mathrm{f}^{-1}(x)\): Domain: \(\underline{x \gt 3}\) or \(\underline{(3, \infty)}\) | B1 |
| (4) |
Notes
M1: Attempt to make \(x\) (or swapped \(y\)) the subject
M1: Makes \(\mathrm{e}^{2x}\) the subject and takes ln of both sides
A1 cao: \(\underline{\tfrac{1}{2}\ln(x - 3)}\) or \(\underline{\ln\sqrt{(x - 3)}}\) or \(\underline{\mathrm{f}^{-1}(y) = \tfrac{1}{2}\ln(y - 3)}\) (see appendix)
B1: Either \(\underline{x \gt 3}\) or \(\underline{(3, \infty)}\) or \(\underline{\text{Domain} \gt 3}\).
| Scheme | Marks |
|---|---|
| \(\mathrm{g}(x) = \ln(x - 1),\ x \in \mathbb{R},\ x \gt 1\) | |
| \(\mathrm{fg}(x) = \mathrm{e}^{2\ln(x - 1)} + 3\ \left\{= (x - 1)^2 + 3\right\}\) | M1 A1 isw |
| \(\mathrm{fg}(x)\): Range: \(\underline{y \gt 3}\) or \(\underline{(3, \infty)}\) | B1 |
| (3) | |
| (15 marks) |
Notes
M1: An attempt to put function g into function f.
A1 isw: \(\mathrm{e}^{2\ln(x - 1)} + 3\) or \((x - 1)^2 + 3\) or \(x^2 - 2x + 4\).
B1: Either \(\underline{y \gt 3}\) or \(\underline{(3, \infty)}\) or \(\underline{\text{Range} \gt 3}\) or \(\underline{\mathrm{fg}(x) \gt 3}\).